Let $G$ be a finite group acting on a topological space $X$. (In my applications, $X$ is the classifying space of a compact Lie group.) Let $H^\ast(-)=H^\ast(-;\mathbb{Z}/2\mathbb{Z})$ denote mod 2 singular cohomology.

By the work of Quillen ("The Spectrum of an Equivariant Cohomology Ring: I"), if $H^\ast(X)$ is a finitely generated $\mathbb{Z}/2\mathbb{Z}$-module, then the equivariant cohomology $$H^\ast_G(X) = H^\ast(EG\times_G X)$$ is a finitely generated as a $\mathbb{Z}/2\mathbb{Z}$-algebra.

My question is, if we only assume that $H^\ast(X)$ is finitely generated *as an algebra*, does the conclusion still hold, ie is the equivariant cohomology $H^\ast_G(X)$ necessarily finitely generated as an algebra?

**More background**

The standard tool for computing equivariant cohomology is the Leray-Serre spectral sequence of the fibration $X\to EG\times_G X \to BG$, which has
$$H^\ast(BG,H^\ast(X)) \Longrightarrow H^\ast(EG\times_G X)$$
*as algebras*.

Now if $G$ acts trivially on $H^\ast(X)$ then the $E_2$-page can be identified with
$$H^\ast(BG)\otimes H^\ast(X)$$
*as algebras*. By a classical result of Evens, the mod 2 cohomology algebra of a finite group is finitely generated. Since the tensor product of finitely generated algebras is again finitely generated, so is this $E_2$-page, ~~and hence so is the equivariant cohomology algebra.~~ **(see the comments made by Algori and Ralph below.)**

However, in general the coefficients in the $E_2$-page are twisted by the action of $G$ on $H^\ast(X)$ (which in my case happens to be non-trivial), and so I can't see how the argument would go.

At the other extreme, if $G$ acts freely on $X$ then we are basically asking if there is a finite covering space $X\to Y$ such that $H^\ast(X)$ is finitely generated as an algebra but $H^\ast(Y)$ is not. Such a beast would give a counter-example.

(I'm also rather sure that if the answer to my question was yes, Quillen would have told us so!)